Official Calculations Discussion Thread

[MonkeyOfLife]

I've been trying to figure out how to find ways to calculate flying slashes, specifically for how strong/energy would you need to send out winds that can destroy something like concrete and for example with __ distance of the slash

Does anyone know how you could calculate something like that?

---

Don't understand it but does this way work/make sense? asked AI

The energy of pressurized air can be calculated using the formula for pressure energy in an incompressible fluid volume, such as a pressurized tank with air. The formula is given by: ΔE = Δp / ρ, where ΔE is the potential energy, Δp is the pressure difference, and ρ is the density of the fluid1. For example, if you have a pressurized tank with air at a pressure of 10 bar (106 Pa) and the density of air is 1.225 kg/m3, the potential pressure energy can be calculated as E = (106 Pa) / (1.225 kg/m3) = 86.53 J/kg. Is there anything else you would like to know?

Or maybe actually this?

The energy of wind can be calculated using the formula for kinetic energy of wind, which is based on the ability to capture the energy contained in air motion1. Wind power quantifies the rate of this kinetic energy extraction1. To estimate wind energy, you can use the formula: E = 0.5 * A * t * ρ * v^3, where E is the wind energy, A is the surface area perpendicular to the wind direction, t is the duration of the wind, ρ is the density of air, and v is the wind speed2. Additionally, wind power is the energy per unit time, so the wind power formula is: P = E / t2.

However, whether or not a wind can destroy objects depends on several factors such as the velocity of wind (actually the square of the velocity), gravity, static friction (the force that keeps you anchored to the ground, along with gravity), drag of the wind pushing on you, air density, your weight, size and center of gravity3. Is there anything else you would like to know?

 

[Hagane_no_Saiyajin]

I'm uncertain about how to calculate this feat, whether it is the speed to perform it or the force that is needed

 

[Shmeatywerbenmanjenson]

https://vsbattles.fandom.com/wiki/User_blog:Arkenis/HXH_-_Netero_Slaps_Pitou is this fine?

I left a comment on the blog

Also did you start getting into HxH?

 

[Arkenis]

I left a comment on the blog

Also did you start getting into HxH?

nah i watched the anime last year, i just never really cared for the profiles till now.

 

[LaserPrecision]

I'm uncertain about how to calculate this feat, whether it is the speed to perform it or the force that is needed

Get the height of the guy being spun, and use an arc calculator. For speed... hmm, maybe looking up the speed at which things blur, and assume they make one full rotation in that timeframe.

Like if you get 50-60 hertz

1/50 = 0.02s

Assuming the dude is like 172cm or smth

(360/360)*2*pi*1.72 = 10.8070787283m

10.8070787283/0.02 = 540.353936415 m/s

That's just an example and makes assumptions, but you get the general idea. For force, you can probably use the regular force method? Mass x Acceleration?

Just assume it took 1 second or something to reach top speed I guess. Or whatever feels appropriate based on panel count and context.

 

[SeijiSetto]

Get the height of the guy being spun, and use an arc calculator. For speed... hmm, maybe looking up the speed at which things blur, and assume they make one full rotation in that timeframe.

Like if you get 50-60 hertz

1/50 = 0.02s

Assuming the dude is like 172cm or smth

(360/360)*2*pi*1.72 = 10.8070787283m

10.8070787283/0.02 = 540.353936415 m/s

That's just an example and makes assumptions, but you get the general idea. For force, you can probably use the regular force method? Mass x Acceleration?

Just assume it took 1 second or something to reach top speed I guess. Or whatever feels appropriate based on panel count and context.

could do centripetal force, (mv^2)/r
said force would be the tension in their arms which they obviously have to withstand or they'd be torn off

 

[LaserPrecision]

could do centripetal force, (mv^2)/r
said force would be the tension in their arms which they obviously have to withstand or they'd be torn off

Yeah, that'd prolly be better. Thank you for the correction.

 

[Oliver_de_jesus]

By the way, how do I calculate that someone would split a board made of a Titanium and carbon alloy in half with one punch?

 

[KLOL506]

By the way, how do I calculate that someone would split a board made of a Titanium and carbon alloy in half with one punch?

Need the crack width, length and board thickness.

 

[LaserPrecision]

Need the crack width, length and board thickness.

I feel the crack width would be rare to find given most feats have the object split apart/start collapsing/moving away from its original position. No?

 

[StretchSebe]

I feel the crack width would be rare to find given most feats have the object split apart/start collapsing/moving away from its original position. No?

Yeah, unless you're able to pixel scale the crack width somehow.

In my experience with similar calculations, I've been corrected that often with human sized objects being split, 0.5 - 1 cm crack width is assumed.

Don't know if that's a hard and fast rule, just anecdote.

 

[LaserPrecision]

Yeah, unless you're able to pixel scale the crack width somehow.

In my experience with similar calculations, I've been corrected that often with human sized objects being split, 0.5 - 1 cm crack width is assumed.

Don't know if that's a hard and fast rule, just anecdote.

I feel it'd prolly be better to just use the width of whatever cut the thing in half. Like if someone punches a wall in half or something, 1cm doesn't seem right at all.

 

[StretchSebe]

I feel it'd prolly be better to just use the width of whatever cut the thing in half. Like if someone punches a wall in half or something, 1cm doesn't seem right at all.

The width of the crack wouldn't be the same as the width/length of the object it's split across. Usually sharing length, but it does have it's own width dependent on the damage done that seems to be considered when splitting calcs are evaluated.

 

[LaserPrecision]

The width of the crack wouldn't be the same as the width/length of the object it's split across. Usually sharing length, but it does have it's own width dependent on the damage done that seems to be considered when splitting calcs are evaluated.

I mean, if you slice a large block clean in half with a chop, that cut left behind would ideally be the width of your hand, no? Otherwise how would the hand glide through the material? Not like the hand can shrink.

 

[StretchSebe]

I mean, if you slice a large block clean in half with a chop, that cut left behind would ideally be the width of your hand, no? Otherwise how would the hand glide through the material? Not like the hand can shrink.

Ah, I should clarify that yeah: Larger objects definitely warrant larger crack widths. Like unless someone is stated to literally split a mountain in half cleanly with a katana or something, there's no way it's 0.5 - 1 cm crack width lol.

As for the exacts, I would again say you'd have to figure it out in reference to something (like a hand) or through pixel scaling.

 

[LaserPrecision]

Ah, I should clarify that yeah: Larger objects definitely warrant larger crack widths. Like unless someone is stated to literally split a mountain in half cleanly with a katana or something, there's no way it's 0.5 - 1 cm crack width lol.

As for the exacts, I would again say you'd have to figure it out in reference to something (like a hand) or through pixel scaling.

I would assume it applies to all objects regardless of size. The split bare minimum needs to be large enough for what cut through it to go through it. The only exception would be shockwave like splits (Like in the mountain katana splitting example where the katana obviously wouldn't be able to cut through the entire thing). Even a hands width on its side (What people would call a "chop") is larger than 1 cm. That's too thin if someone cut through the material with their hand. They should use the width of someone's hand (or measure their own ig). Which is like 1-2 inches or smth.

 

[StretchSebe]

(Like in the mountain katana splitting example where the katana obviously wouldn't be able to cut through the entire thing).

Beeg katana.

Even a hands width on its side (What people would call a "chop") is larger than 1 cm. That's too thin if someone cut through the material with their hand. They should use the width of someone's hand (or measure their own ig). Which is like 1-2 inches or smth.

Again, if you have something to reference, like a persons hand chop, or even something thick like a club, I'd assume it'd be fine to use that width for crack width.

I'm no authority on this; just providing my own experience with what seemed to be the norm when you don't have much to pixel scale or reference.

 

[Hagane_no_Saiyajin]

What's the AP needed to do this:

 

[Arkenis]

What's the AP needed to do this:

is this Dai Dark or Dorohedoro?

 

[Hagane_no_Saiyajin]

is this Dai Dark or Dorohedoro?

Dorohedoro

 

[SeijiSetto]

i am looking for fragmentation values for tempered glass, but i cannot find any. there's a character who punches a hollow cylinder of it hard enough to make it shatter into large chunks.
according to THIS blog, fragmentation comes from the shear strength of the material, but lo and behold - nobody's got anything.
the closest i found was a couple sources saying its COMPRESSIVE strength is 1000 MPa and this source but it's pretty incomplete and iffy to use.

 

[DontTalkDT]

Primarily used for throwing shit.


No, that shit requires angular momentum, though from what I recall, you aren't allowed to calculate force from swinging massive objects like bats according to this.

However, this seems very vague at first glance. So I will ask @DontTalkDT for real this time.

@DontTalkDT Can you calculate force for swinging large objects like bats, like in this case? Ugarik used your reply to justify not doing so, but should that really be the case? Just asking.

Well

The best you can probably do is say that it would be at least as much as in the constant acceleration case. (aka approximate via the low end)

If you do it via the low in regard to acceleration I think it's probably ok.

The point I was trying to make in the reply is that you don't calculate KE and Force with the same assumption, as what is a low end to one is a high-end to the other. Sorry if I wasn't really clear.

 

[KLOL506]

Well

If you do it via the low in regard to acceleration I think it's probably ok.

The point I was trying to make in the reply is that you don't calculate KE and Force with the same assumption, as what is a low end to one is a high-end to the other. Sorry if I wasn't really clear.

So uh... me dumb, could you suggest the appropriate formulas to calculate force outta this thing? We already got the KE so uh...

And if you would be so kind, would you be able to calculate force for this to show an example of how we should tackle these kinda feats?

 

[SeijiSetto]

So uh... me dumb, could you suggest the appropriate formulas to calculate force outta this thing? We already got the KE so uh...

And if you would be so kind, would you be able to calculate force for this to show an example of how we should tackle these kinda feats?

could you not just divide the KE by the length of the swing? basically equating kinetic energy (1/2mv^2) to work done (Fd) and then dividing by d to get F in newtons.

i'm pretty sure similar things are allowed when throwing things, get the KE of the thrown object and divide it by the distance of the arm-swing needed to throw it

 

[DontTalkDT]

So uh... me dumb, could you suggest the appropriate formulas to calculate force outta this thing? We already got the KE so uh...

And if you would be so kind, would you be able to calculate force for this to show an example of how we should tackle these kinda feats?

could you not just divide the KE by the length of the swing? basically equating kinetic energy (1/2mv^2) to work done (Fd) and then dividing by d to get F in newtons.

i'm pretty sure similar things are allowed when throwing things, get the KE of the thrown object and divide it by the distance of the arm-swing needed to throw it

Essentially this. If you assume constant acceleration force can be calculated by dividing KE through distance.
Otherwise, you could try to measure the actual acceleration, but that's a lot more complicated.

 

[KLOL506]

Essentially this. If you assume constant acceleration force can be calculated by dividing KE through distance.
Otherwise, you could try to measure the actual acceleration, but that's a lot more complicated.

Divide KE by distance travelled huh.

So the sword length as per this is 36.85 m.

Times 2 radians is 73.7 meters travelled.

W = Fs

F = W/s.

W = 6.6667e+13 J

F = 6.6667e+13/73.7 = 904572591588 N or 9.2209234616513761468e+10 kgf (Class G)

I get it right?

 

[BreezeHM]

If it was stated someone has the power to destroy a star, do we scale them to baseline star level? Or do we have to do a calculation?

 

[Stryker861]

If it was stated someone has the power to destroy a star, do we scale them to baseline star level? Or do we have to do a calculation?

Baseline

 

[Majinere566]

Question.

If a 1,000km structure isn't anywhere near outerspace of a planet. Is that enough proof of the planet being bigger than Earth?

And if so how do I go about finding it's size?

 

[SeijiSetto]

Divide KE by distance travelled huh.

So the sword length as per this is 36.85 m.

Times 2 radians is 73.7 meters travelled.

W = Fs

F = W/s.

W = 6.6667e+13 J

F = 6.6667e+13/73.7 = 904572591588 N or 9.2209234616513761468e+10 kgf (Class G)

I get it right?

eyup

 

[Majinere566]

Question.

If a 1,000km structure isn't anywhere near outerspace of a planet. Is that enough proof of the planet being bigger than Earth?

And if so how do I go about finding it's size?

Bump

 

[Arkenis]

Bump

Yeah. the spheres of our planet only start reaching that height just before the exosphere comes into play.

 

[Majinere566]

Yeah. the spheres of our planet only start reaching that height just before the exosphere comes into play.

Alright, is there anyway that I can go about calcing it's size?

 

[Arkenis]

Alright, is there anyway that I can go about calcing it's size?

Uh would depend on the way the structure is shown so first find a good pic that shows the structure and the world.

 

[Majinere566]

Uh would depend on the way the structure is shown so first find a good pic that shows the structure and the world.

well there's no picture of said world however there's a clear picture of the structure.

 

[Arkenis]

well there's no picture of said world however there's a clear picture of the structure.

Send

 

[Majinere566]

Send

Alright looking back at it, it's not the clearest, but there's a clear picture of its head and I found this scan of its head silhouette.

Scan 1

Scan 2

 

[Hagane_no_Saiyajin]

What's the AP needed to do this:

boop

 

[SeijiSetto]

i am looking for fragmentation values for tempered glass, but i cannot find any. there's a character who punches a hollow cylinder of it hard enough to make it shatter into large chunks.
according to THIS blog, fragmentation comes from the shear strength of the material, but lo and behold - nobody's got anything.
the closest i found was a couple sources saying its COMPRESSIVE strength is 1000 MPa and this source but it's pretty incomplete and iffy to use.

bump

also when calculating the volume of robots and other mechanical things, what's the go-to hollowness value? as in, the way that people will apply 80 or 90% hollowness when calculating the volume or mass of a building destroyed.
i THINK i've seen 70% mentioned before but i have no source for that. currently calculating a feat where a character heavily damages one with a single punch, so i need to know the actual volume of steel broken.

 

[KLOL506]

bump

also when calculating the volume of robots and other mechanical things, what's the go-to hollowness value? as in, the way that people will apply 80 or 90% hollowness when calculating the volume or mass of a building destroyed.
i THINK i've seen 70% mentioned before but i have no source for that. currently calculating a feat where a character heavily damages one with a single punch, so i need to know the actual volume of steel broken.

Robots and other mechanical things usually go for 50% from my knowledge